Question

think about the expansion for the binomial $(2x - y^2)^6$.
the first term of this binomial is $2x$.
the second term of this binomial is $-y^2$.
what is the correct formula for the fifth term in this binomial expansion?
options:
$ _6c_5(2x)(-y^2)^5 $
$ _6c_4(2x)^4(-y^2)^2 $
$ _6c_4(2x)^4(-y^2)^4 $

Explanation:

Step1: Recall Binomial Theorem

The binomial theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\). For the binomial \((2x-y^{2})^{6}\), we have \(a = 2x\), \(b=-y^{2}\), and \(n = 6\).

Step2: Determine the term number

We want the fifth term. In the binomial expansion, the \(r\)-th term (starting from \(r = 1\)) corresponds to \(k=r - 1\) in the summation. So for the fifth term, \(r = 5\), which means \(k=5 - 1=4\).

Step3: Apply the binomial formula

Substitute \(n = 6\), \(k = 4\), \(a = 2x\), and \(b=-y^{2}\) into the binomial formula \(\binom{n}{k}a^{n - k}b^{k}\). We get \(\binom{6}{4}(2x)^{6 - 4}(-y^{2})^{4}\)? Wait, no, wait. Wait, \(n-k=6 - 4 = 2\)? Wait, no, wait, let's re - check. Wait, the general term is \(\binom{n}{k}a^{n - k}b^{k}\), where \(k\) starts from \(0\). So the first term is when \(k = 0\), the second when \(k = 1\),..., the fifth term is when \(k=4\) (since \(k = 0\) is term 1, \(k = 1\) term 2, \(k = 2\) term 3, \(k = 3\) term 4, \(k = 4\) term 5). So \(n=6\), \(k = 4\), \(a = 2x\), \(b=-y^{2}\). Then the term is \(\binom{6}{4}(2x)^{6 - 4}(-y^{2})^{4}\)? Wait, no, the options: let's look at the options. Wait, the options are:
1. \(_6C_5(2x)(-y^{2})^{5}\)
2. \(_6C_4(2x)^{4}(-y^{2})^{2}\)
3. \(_6C_4(2x)^{2}(-y^{2})^{4}\) Wait, no, let's re - calculate \(n-k\) when \(k = 4\): \(n-k=6 - 4=2\), so \(a^{n - k}=(2x)^{2}\), and \(b^{k}=(-y^{2})^{4}\)? Wait, no, maybe I made a mistake. Wait, the binomial is \((2x - y^{2})^{6}=(2x+(-y^{2}))^{6}\). The general term is \(\binom{6}{k}(2x)^{6 - k}(-y^{2})^{k}\). For the fifth term, since the first term is \(k = 0\) (term 1), term 2: \(k = 1\), term 3: \(k = 2\), term 4: \(k = 3\), term 5: \(k = 4\). So when \(k = 4\), the term is \(\binom{6}{4}(2x)^{6 - 4}(-y^{2})^{4}\)? Wait, no, \(6-4 = 2\), so \((2x)^{2}\), and \((-y^{2})^{4}\). But looking at the options, the second option is \(_6C_4(2x)^{4}(-y^{2})^{2}\), the third is \(_6C_4(2x)^{2}(-y^{2})^{4}\)? Wait, maybe I mixed up \(k\) and \(n - k\). Wait, another way: the \(r\)-th term in the expansion of \((a + b)^n\) is given by \(\binom{n}{r - 1}a^{n-(r - 1)}b^{r - 1}\). So for the fifth term, \(r = 5\), so \(\binom{6}{5 - 1}(2x)^{6-(5 - 1)}(-y^{2})^{5 - 1}=\binom{6}{4}(2x)^{2}(-y^{2})^{4}\)? Wait, no, \(6-(5 - 1)=6 - 4 = 2\), \(5 - 1 = 4\). But the second option is \(_6C_4(2x)^{4}(-y^{2})^{2}\), which would be when \(k = 2\) (term 3). Wait, maybe the question has a typo, or I made a mistake. Wait, let's check the exponent of \(2x\) and \(-y^{2}\) in each option:

Option 1: \(_6C_5(2x)(-y^{2})^{5}\): \(k = 5\), which is term 6 (since \(k = 0\) is term 1, \(k = 5\) is term 6)
Option 2: \(_6C_4(2x)^{4}(-y^{2})^{2}\): \(k = 2\) (since \(b\) has exponent 2), so \(n - k=6 - 2 = 4\), so term number is \(k + 1=3\) (term 3)
Option 3: Wait, the third option (the middle one) is \(_6C_4(2x)^{2}(-y^{2})^{4}\): \(k = 4\), so term number is \(k + 1=5\) (since \(k = 0\) is term 1, \(k = 4\) is term 5). Wait, but let's re - check the binomial theorem. The general term is \(\binom{n}{k}a^{n - k}b^{k}\), where \(k\) starts at 0. So term 1: \(k = 0\), term 2: \(k = 1\), term 3: \(k = 2\), term 4: \(k = 3\), term 5: \(k = 4\). So for term 5, \(k = 4\), so \(a^{n - k}=(2x)^{6 - 4}=(2x)^{2}\), \(b^{k}=(-y^{2})^{4}\), and \(\binom{n}{k}=\binom{6}{4}\). So the term is \(\binom{6}{4}(2x)^{2}(-y^{2})^{4}\). But looking at the options, the second option (the middle box) is \(_6C_4(2x)^{4}(-y^{2})^{2}\), which is for \(k = 2\) (term 3), and the first option is for \(k = 5\) (term 6), and the third option (the bottom one) is \(_6C_4(2x)^{2}(-y^{2})^{4}\)? Wait, maybe the user's image has the options as:

First box: \(_6C_5(2x)(-y^{2})^{5}\)

Second box: \(_6C_4(2x)^{4}(-y^{2})^{2}\)

Third box: \(_6C_4(2x)^{2}(-y^{2})^{4}\)

Wait, no, let's re - calculate the exponent correctly. The binomial is \((2x - y^{2})^{6}\). The fifth term:

Term 1: \(k = 0\): \(\binom{6}{0}(2x)^{6}(-y^{2})^{0}\)

Term 2: \(k = 1\): \(\binom{6}{1}(2x)^{5}(-y^{2})^{1}\)

Term 3: \(k = 2\): \(\binom{6}{2}(2x)^{4}(-y^{2})^{2}\)

Term 4: \(k = 3\): \(\binom{6}{3}(2x)^{3}(-y^{2})^{3}\)

Term 5: \(k = 4\): \(\binom{6}{4}(2x)^{2}(-y^{2})^{4}\)

Ah! So the fifth term is \(\binom{6}{4}(2x)^{2}(-y^{2})^{4}\), which is the third option (the bottom box). Wait, but the options given in the image:

First box: \(_6C_5(2x)(-y^{2})^{5}\) (this is term 6, since \(k = 5\), term \(5 + 1=6\))

Second box: \(_6C_4(2x)^{4}(-y^{2})^{2}\) (this is term 3, since \(k = 2\), term \(2+1 = 3\))

Third box: \(_6C_4(2x)^{2}(-y^{2})^{4}\) (this is term 5, since \(k = 4\), term \(4 + 1=5\))

So the correct formula for the fifth term is \(_6C_4(2x)^{2}(-y^{2})^{4}\), which is the third option (the bottom box). But let's check the options again. Wait, maybe I misread the exponents. Wait, the second option is \(_6C_4(2x)^{4}(-y^{2})^{2}\), which is term 3, and the third option is \(_6C_4(2x)^{2}(-y^{2})^{4}\), which is term 5. So the correct answer is the third option, \(_6C_4(2x)^{2}(-y^{2})^{4}\) (the bottom box). But let's confirm with the binomial theorem formula. The general term \(T_{k + 1}=\binom{n}{k}a^{n - k}b^{k}\). For the fifth term, \(k+1 = 5\), so \(k = 4\). Then \(T_{5}=\binom{6}{4}(2x)^{6 - 4}(-y^{2})^{4}=\binom{6}{4}(2x)^{2}(-y^{2})^{4}\), which matches the third option.

Answer:

\(_6C_4(2x)^{2}(-y^{2})^{4}\) (the bottom box option)